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[Hint :Sx – 1 = S49 – Sx]

Answer :

Given,

Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

So,

First term, a = 1

Common difference, d=1

Let us say the number of \(x^{th}\) houses can be represented as;

Sum of nth term of AP = \({{n} \over {2}}[2a + (n - 1)d]\)

Sum of number of houses beyond x house = Sx-1

= \({{x - 1} \over {2}}[2(1) + (x -1-1)1]\)

= \({{x - 1} \over {2}} [2+x-2]\)

= \({{x(x - 1)} \over {2}}\)………………………………(i)

By the given condition, we can write,

\(S_{49} – Sx = ({{49} \over {2}}[2.1+(49-1)1])–({{x} \over {2}}[2.1+(x-1)1])\)

= \(25(49) – {{x( x + 1)} \over {2}} \)………………………………….(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other;

Therefore,

\({{x(x - 1)} \over {2}} = 25(49) – {{x ( x - 1)} \over {2}}\)

\(x = ±35\)

As we know, the number of house cannot be an a negative number. Hence, the value of x is 35

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- A ladder has rungs 25 cm apart.. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \({2} {\dfrac{1}{2}}\) apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = -250/25 ].
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